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2 Process DirectionNo apparatus can operate in such a way that its only effect (in system & surroundings) is to convert heat absorbed by a system completely into work done by the systemIt is impossible by a cyclic process to convert the heat absorbed by a system completely into work done by the system.No process is possible which consist solely in the transfer of heat from one temperature level to a higher one.Partial conversion of heat to into work is the basis for nearly all commercial production of power.

3 Heat Engine Heat engines produce work from heat in a cyclic processEssential to all heat-engine cycles are:Absorption of heat into the system at a high temperatureRejection of heat to surroundings at a lower temperatureProduction of work

4 Conversion of heat to work always accompanied by heat release to surroundings

5 Heat Engine Thermal efficiency:For  to be unity, QC must be zero. No engine has ever been built for which this is true; some heat is always rejected.

6 Carnot’s Theorem & Carnot EngineCarnot engine operates between two heat reservoirs in such a way that all heat absorbed is absorbed at the constant temperature of the hot reservoir and all heat rejected at the constant temperature of the cold reservoirFor two given heat reservoir, no engine can have a thermal efficiency higher than that of a Carnot engine.

7 Carnot CycleStep1: A system at TC undergoes a reversible adiabatic process that causes its temperature rise to TH Step 2: The system maintains contact with the hot reservoir TH, and undergoes a reversible isothermal process during which heat QH is absorbed from the hot reservoir Step 3: The system undergoes a reversible adiabatic process in the opposite direction of step 1 that brings its temperature back to that of the cold reservoir at TC.Step 4: The system maintains contact with the reservoir at TC, and undergoes reversible isothermal process in the opposite direction of the step 2 that returns it to its initial state with heat rejection of QC to the cold reservoir

10 EntropyOur analysis showed that or The equation suggests the existence of a property whose changes are given by the quantities Q/T When the isotermal steps are infinitesimal, the heat quantities become dQ: or Thus the quantities dQrev/T sum to zero for the arbitrary cycle, exhibiting the characteristic of a property. We call this property as entropy, S.There exists a property called entropy S, which is an intrinsic property of a system, functionally related to the measurable coordinates which characterize the system

11 Characteristic of EntropyEntropy owes its existence to the second law, from which it arises in much the same way as internal energy does from the first law.The change in entropy of any system undergoing a finite reversible process isWhen a system undergoes an irreversible process between two states, the entropy change of the system is evaluated to an arbitrary chosen reversible process that accomplishes the same change of state as the actual process. The integration is not carried out for the irreversible path.

12 Characteristic of EntropySince entropy is a state function, the entropy changes of the irreversible and reversible processes are identical.The entropy change of a system caused by the transfer of heat can always be calculated by dQ/T whether the heat transfer is accomplished reversibly or irreversibly.When a process is irreversible on account of finite differences in other driving forces, such as pressure, the entropy changes is not caused solely by heat transfer, and for its calculations one must devise a reversible means of accomplishing the same change of state.

13 Calculation of EntropyFor one mole of fluid undergoing a mechanically reversible process in a closed system, using the first law, the defining equation for enthalpy one find: For an ideal gas: Although derived for a mechanically reversible process, this equation is a general equation for the calculation of entropy changes, since it relates properties only and independent of process causing the change of state.

15 Problem 5.6A quantity of an ideal gas, Cp = 7/2 R at 20oC & 1 bar and having a volume of 70 m3 is heated at constant pressure to 25oC by the transfer of heat from a heat reservoir at 40oC. Calculate the heat transfer to the gas, the entropy change of the heat reservoir, the entropy change of the gas, and DStotal.

16 Problem 5.7A rigid vessel of 0.05 m3 volume contains an ideal gas, Cv = 5/2 R, at 500 K and 1 bar.If heat in the amount of J is transferred to the gas, determine its entropy changeIf the vessel is fitted with a stirrer that is rotated by a shaft so that work on the amount of J is done on the gas, what is the entropy change of the gas if the process is adiabatic? What is DStotal?

17 Example 5.3Methane gas at 550 K & 5 bar undergoes a reversible adiabatic expansion to 1 bar. Assuming methane to be an ideal gas at these conditions, determine its final temperature.

19 Heat Engine Thermal efficiency:For  to be unity, QC must be zero. No engine has ever been built for which this is true; some heat is always rejected.

20 Mathematical Statement of The Second LawThis mathematical statement of the second law affirms that every process proceeds in such a direction that the total entropy change associated with it is positive.The limiting value of zero being attained only by a reversible process.No process is possible for which the total entropy decreases.

21 Thermodynamic EfficiencyIn a process producing work, there is an absolute maximum work attainable which is accomplished by completely reversible process. For irreversible process, Wactual produced < Wideal In a process requiring work, there is an absolute minimum amount of work required which is accomplished by completely reversible process. For irreversible process, Wractual required > Wideal

23 Problem 5.13A reversible cycle executed by 1 mol of an ideal gas for which Cp = 5/2 R consist of the following processes:Starting at 600 K & 2 bar, the gas is cooled at constant pressure to 300 KFrom 300 K & 2 bar, the gas is compressed isothermally to 4 barThe gas returns to its initial state along a path for which the product PT is constant.What is the thermal efficiency of the cycle?

24 Example 5.4A 40-kg steel casting (Cp = 0.5 kJ/(kg.K) at a temperature of 450oC is quenched in 150 kg of oil (Cp = 2.5 kJ/(kg.K) at 25oC. If there are no heat losses, what is the change of entropy of:The castingThe oilBoth considered together

25 Problem 5.11A piston/cylinder device contains 5 mol of an ideal gas, Cp = 5/2 R, at 20oC & 1 bar. The gas is compressed reversibly and adiabatically to 10 bar, where the piston is locked in position. The cylinder is then brought into thermal contact with a heat reservoir at 20oC, and heat transfer continues until the gas also reaches this temperature. Determine the entropy changes of the gas, the reservoir, and DStotal

26 IrreversibilityOne mole of an ideal gas, Cp = 7/2 R is compressed adiabatically in a piston/cylinder device from 2 bar & 25oC to 7 bar. The process is irreversible and requires 35% more work than a reversible adiabatic compression from the same initial state to the same final pressure. What is the entropy change of the gas?

27 Classical Lost Work & Process AnalysisThe ideal work is the maximum amount of work which can be done by the process by operating reversibly within the system and by transferring heat between the system and the surroundings reversibly. The lost work: For processes containing several units, lost-work calculations can be made for each unit and summed to determine the overall value.

28 Example 4.7Steam enters a turbine at 1.5 MPa & 500oC and exhausts at 0.1 MPa. The turbine delivers 85% of the shaft work of a reversible-adiabatic turbine although it is neither reversible nor adiabatic. Heat losses to the surroundings at 20oC are 9 kJ/kg steam. Determine temperature, entropy change of the steam leaving the turbine, & the lost work

29 Example 4.8Assume that 5000 kg/h of oil with a heat capacity of 3.2 kJ/kg.K is to be cooled from 220 to 40oC, using a large quantity of water which can be assumed to be at a constant temperature of 30oC. Determine the lost work in the process and the thermodynamic efficiency of the process.

30 Example 4.9Assume that 100 kg of methane gas/h is adiabatically compressed from 0.5 MPa & 300 K to 3.0 MPa & 500 K after which it is cooled isobarically to 300 K by a large amount of water available at 290 K. Determine the efficiency of the compressor. If the surroundings are assumed to be at 290 K, do a thermodynamic analysis of the process. Assume ideal gas. Cp = J/mol.K Cv = J/mol.K

35 Problem 5.8An ideal gas, Cp = 7/2 R, is heated in a steady-flow heat exchanger from 20oC to 100oC by another stream of the same ideal gas which enters at 180oC. The flow rates of the two streams are the same, and heat losses from the exchanger are negligible.Calculate the molar entropy changes of the two gas streams for both parallel and counter-current flow in the exchangerWhat is DStotal in each case?